How to Evaluate the Sum of Levi-Civita Symbols?

[sleventh]

Symbolic Linear algebra concept

Homework Statement

Evaluate the sum \Sigmak ( \epsilon ijk \epsilon lmk (which contains three terms) by considering the result for all possible combinations of i,j,l,m, that is a) i=j b) i=l c) i=m d) j=l e)j=m f)l=m g) i\neql or m h) j\neql or m

show that

\Sigmak ( \epsilon ijk \epsilon lmk = \delta il \delta jm - \delta I am \delta jl

and use this result to prove
A X (B X C)=(A \bullet C)B-(A \bullet B)C

Homework Equations

The Attempt at a Solution

i have been working on this problem for over three hours and have also gone to my professor which was of little help, the work I have done would most likely only serve to confuse. I would be EXTREMELY grateful to anyone who would be able to help me understand this problem. I understand for part a) the permittivity will always equal 0 for the first Levi-Civita symbol making the entire sum zero. For question like be I am lost as to how you could know which values would be +1 or -1 or 0 out of the 81 possible combinations. As for the proofs I am lost on the first and the second I am having a hard time manipulating the concept A X B = \Sigma j, k \epsilon ijk AjBk . Thank you very much to anyone who can shed the slightest of light.
yours truly
sleventh

 

[tiny-tim]

Hi sleventh!

(have a sigma: ∑ and a delta: δ and an epsilon: ε and try using the X₂ tag just above the Reply box )

Evaluate the sum \Sigmak ( \epsilon ijk \epsilon lmk (which contains three terms) by considering the result for all possible combinations of i,j,l,m, that is a) i=j b) i=l c) i=m d) j=l e)j=m f)l=m g) i\neql or m h) j\neql or m

show that

\Sigmak ( \epsilon ijk \epsilon lmk = \delta il \delta jm - \delta I am \delta jl

and use this result to prove
A X (B X C)=(A \bullet C)B-(A \bullet B)C
…
I understand for part a) the permittivity will always equal 0 for the first Levi-Civita symbol making the entire sum zero. For question like be I am lost as to how you could know which values would be +1 or -1 or 0 out of the 81 possible combinations.

(what does permittivity have to do with it? )

No, there aren't 81 …

for example, for b) i = l, put i = l = 1, then there are only nine combinations for j and k, and most of them are zero!

Try again.

As for the proofs I am lost on the first and the second I am having a hard time manipulating the concept A X B = \Sigma j, k \epsilon ijk AjBk

(you mean (AxB)_i = …)

Just look at the "matrix" definition of AxB (where e_x e_y and e_z are the basis):

ex  ey  ez
Ax  Ay  Az
Bx  By  BZ

 

[sleventh]

thank you very much tiny tim, It was silly not to set the matrix into its components, i was only oriented to the summation equation. i still am lost on LaTeX Code: \\Sigma k ( LaTeX Code: \\epsilon ijk LaTeX Code: \\epsilon lmk = LaTeX Code: \\delta il LaTeX Code: \\delta jm - LaTeX Code: \\delta I am LaTeX Code: \\delta jl
thank you very much, you helped clear many things up.

 

[tiny-tim]

hmm … your LaTeX looks a bit weird …

the PF version is very simple, it doesn't use \\, and for subscripts just type _{ijk}

(and use \cdot not \bullet)

For more details, see http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000"

(but it's easier on the server if you use standard Unicode symbols and the X² and X₂ tags instead )

 

[lurflurf]

No, there aren't 81 …

If we have n=3 there are 81 as 3^4=81.
The fancy way to do this would be to observe
sum epsilon(ijk)epsilon(lmk)
is isotropic so can be expressed in the form
A*delta(ij)delta(lm)+B*delta(il)delta(jm)+C*delta(im)delta(jl)
since
{delta(ij)delta(lm),delta(il)delta(jm),delta(im)delta(jl)}
is a basis for the isotropic 4th order 3 spance tensors

The tedious method (though not that tedious) as you mentioned is to group 81 componets into groups based on there possible values -1,0,1.

 

[sleventh]

thank you lurflurf, i have been unaware how to put the epsilon notation into delta form, however in retrospect it doesn't seem to have been to difficult. Because of this i did have to sort the 81 possibilities.
I am though lost on how to actually prove the sum of the two epsilons to equal the Kronecker delta equation. If i figure this out i will post the solution.
thanks to everyone who helped
sleventh

 

[lurflurf]

I am though lost on how to actually prove the sum of the two epsilons to equal the Kronecker delta equation. If i figure this out i will post the solution.
thanks to everyone who helped
sleventh

If you take the equation as given, it suffices to show the two sides agree for each of the 81 components. Do you mean how to derive the expression?
from
sum epsilon(ijk)epsilon(lmk)=A*delta(ij)delta(lm)+B*delta(il)delta(jm)+C*delta( im)delta(jl)
consider for example
i=1,j=1,l=2,m=2
sum epsilon(11k)epsilon(22k)=A*delta(11)delta(22)+B*delta(12)delta(12)+C*delta( 12)delta(12)
0=A
i=1,j=2,l=1,m=2
sum epsilon(12k)epsilon(12k)=A*delta(12)delta(12)+B*delta(11)delta(22)+C*delta( 12)delta(12)
1=B
likewise for C